By consistency of the covariance estimates, \(\hat{\mathcal {Q}} \rightarrow \mathcal {Q}\)  as \(N \rightarrow \infty \)  in probability (actually, even almost surely). Since the function \(\Phi \)  is continuous, we get: \(\Phi (\hat{\mathcal {Q}} \rightarrow \Phi (\mathcal {Q})\)  and therefore \(\Phi (\hat{\mathcal {Q}})\otimes \Phi (\hat{\mathcal {Q}}) \rightarrow \Phi (\mathcal {Q})\otimes \Phi (\mathcal {Q})\) . Applying Slutsky's theorem [1]} for random vectors to (REF ), we get:
\(N^{1/2}\Phi (\hat{\mathcal {Q}})\otimes \Phi (\hat{\mathcal {Q}})\left[\hat{\mathbb {Q}}_1, \ldots , \hat{\mathbb {Q}}_K\right] \Rightarrow \Phi (\mathcal {Q})\otimes \Phi (\mathcal {Q})\left[Y_1, \ldots , Y_K\right] \stackrel{d}{=} \left[Z_1, \ldots , Z_K\right].\) 

for i.i.d. \(Y_k \sim \mathcal {N}_{m^2}(\mathbf {0}, \mathbf {Q})\)  and i.i.d. \(Z_k \sim \mathcal {N}_{m^2}(\mathbf {0}, \mathbf {Q})\) . This completes the proof of (REF ). The proof of (REF ) is similar.
